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stealth61 [152]
2 years ago
6

Dhorpatan hunting reserve doesn't affect rare animals.​

Physics
1 answer:
Eddi Din [679]2 years ago
4 0

Answer:

Pheasants and partridge are common and their viable population in the reserve permits controlled hunting. Endangered animals in the reserve include Musk deer, Wolf, Red panda, Cheer pheasant and Danphe. The hunting license is issued by the Department of National Parks and Wildlife Conservation.





HOPE THIS HELPS, HAVE A GREAT DAY!!

Explanation:

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63 J of heat are added to a closed system. The initial energy of the system is 58J, and the final internal energy is 93J. How mu
Nesterboy [21]

ΔU = Q + W  

  • heat in, Q +  
  • heat out, Q -  
  • does work , W -  
  • work in, W +

93-58 = 63 + W

35 = 63 + W

W = - 28 J (does work/being done by system)

7 0
2 years ago
How many foot-pounds of work does it take to throw a baseball 90 mph? a baseball weighs 5 oz, or 0.3125 lb?
NARA [144]

Kinetic energy of the ball is (mv²) / 2, where m is the mass and v is the velocity

So plugging in the mass and the velocity into the kinetic energy expression, you get:

Kinetic energy of the ball = (mv²) / 2

(0.3125/32) times (132)² divided by 2 = 85 ft-lbs


Kinetic energy of the ball = 85 ft-lbs


5 0
3 years ago
Which one is it ASAP
Ne4ueva [31]
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7 0
3 years ago
Snowboarder Jump—Energy and Momentum
soldier1979 [14.2K]

Answer:

THE ANSWER TERMS ARE DEFINED BLOW:-

Explanation:

MOMENTUM- IT IS THE ABILITY TO INCREASE OR DEVELOP CONSTANT FORCE.

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POTENTIAL ENERGY:- IT IS THE ENERGY THAT A PARTICLE POSSES WHEN IT ACTUALLY IS IN RESTING STATE.

IN THIS ACIVITY THE SNOWBOARDER IS IN THE MOTION STATE THEREFORE HE POSSES KINETIC ENERGY AND TO MAINTAIN THAT KINEITC ENERG FOR A PERIOD OF TIME,MOMENTUM PLAYS IT'S ROLE.

4 0
3 years ago
At 900.0 K, the equilibrium constant (Kp) for the following reaction is 0.345. 2SO2(g)+O2(g)→2SO3(g) At equilibrium, the partial
elena55 [62]

Answer : The partial pressure of SO_3 is, 67.009 atm

Solution :  Given,

Partial pressure of SO_2 at equilibrium = 30.6 atm

Partial pressure of O_2 at equilibrium = 13.9 atm

Equilibrium constant = K_p=0.345

The given balanced equilibrium reaction is,

2SO_2(g)+O_2(g)\rightleftharpoons 2SO_3(g)

The expression of K_p will be,

K_p=\frac{(p_{SO_3})^2}{(p_{SO_2})^2\times (p_{O_2})}

Now put all the values of partial pressure, we get

0.345=\frac{(p_{SO_3})^2}{(30.6)^2\times (13.9)}

p_{SO_3}=67.009atm

Therefore, the partial pressure of SO_3 is, 67.009 atm

6 0
2 years ago
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